Set up, but do not evaluate, an integral for the area of the surface obtained by rotating the curve about the given axis. $y=e^{x}, 1 \leq y \leq 9 ;$ y-axis
A) $\int_{1}^{9} 2 \pi x \sqrt{1+e^{2 x}} d x$
B) $\int_{1}^{9} 2 \pi x e^{9 x} \sqrt{1+e^{9 x}} d x$
C) $\int_{0}^{\ln 9} 2 \pi x \sqrt{1+e^{2 x}} d x$
D) $\int_{1}^{9} 2 \pi \sqrt{1+e^{2 x}} d x$
E) $\int_{0}^{\ln 9} 2 \pi x \sqrt{1+e^{9 x}} d x$

Question

Quizplus · Accepted Answer

We are rotating the curve $y=e^{x}$ about the $y$-axis, so we need to use the formula $A = \int_{a}^{b} 2\pi x f(x) \sqrt{1+(f'(x))^2}dx$. In this case, $a=1$, $b=9$, $f(x) = e^x$, and we are rotating about the $y$-axis so $x=y$. Thus, the integral we need to set up is:

$$A = \int_{1}^{9} 2\pi y \cdot e^{\ln y} \cdot \sqrt{1+(\frac{d}{dx}e^x)^2} dy = \int_{1}^{9} 2\pi y \cdot e^{\ln y} \cdot \sqrt{1+e^{2\ln y}} dy $$

Simplifying under the radical, we get:

$$A = \int_{1}^{9} 2\pi y \cdot e^{\ln y} \cdot \sqrt{1+y^2} dy $$

Making the substitution $u = \ln y$, we get $y = e^u$ and $dy = e^u du$. Also, when $y=1$, $u = \ln 1 = 0$ and when $y=9$, $u = \ln 9$. Therefore, the integral becomes:

$$A = \int_{0}^{\ln 9} 2\pi e^u \cdot e^{u} \cdot \sqrt{1+e^{2u}} e^u du = \int_{0}^{\ln 9} 2\pi e^{3u} \sqrt{1+e^{2u}} du $$

Which matches the integrand in choice (C).

Set Up, but Do Not Evaluate, an Integral for the Area